Answer
$$\frac{{dy}}{{dx}} = \frac{2}{{5\left( {{x^2} - 1} \right)}}\root 5 \of {\frac{{x - 1}}{{x + 1}}} $$
Work Step by Step
$$\eqalign{
& y = \root 5 \of {\frac{{x - 1}}{{x + 1}}} \cr
& {\text{using radical properties}} \cr
& y = \frac{{\root 5 \of {x - 1} }}{{\root 5 \of {x + 1} }} \cr
& y = \frac{{{{\left( {x - 1} \right)}^{1/5}}}}{{{{\left( {x + 1} \right)}^{1/5}}}} \cr
& {\text{taking the natural logarithm on both sides of the equation}} \cr
& \ln y = \ln \left[ {\frac{{{{\left( {x - 1} \right)}^{1/5}}}}{{{{\left( {x + 1} \right)}^{1/5}}}}} \right] \cr
& {\text{quotient rule of logarithms}} \cr
& \ln y = \ln {\left( {x - 1} \right)^{1/5}} - \ln {\left( {x + 1} \right)^{1/5}} \cr
& {\text{power rule of logarithms}} \cr
& \ln y = \frac{1}{5}\ln \left( {x - 1} \right) - \frac{1}{5}\ln \left( {x + 1} \right) \cr
& {\text{Differentiate both sides}} \cr
& \frac{d}{{dx}}\left( {\ln y} \right) = \frac{d}{{dx}}\left( {\frac{1}{5}\ln \left( {x - 1} \right)} \right) - \frac{d}{{dx}}\left[ {\frac{1}{5}\ln \left( {x + 1} \right)} \right] \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = \frac{1}{5}\left( {\frac{1}{{x - 1}}} \right) - \frac{1}{5}\left( {\frac{1}{{x + 1}}} \right) \cr
& {\text{Solve for dy/dx}} \cr
& \frac{{dy}}{{dx}} = \frac{y}{5}\left( {\frac{1}{{x - 1}} - \frac{1}{{x + 1}}} \right) \cr
& {\text{substituting }}y = \root 5 \of {\frac{{x - 1}}{{x + 1}}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{5}\root 5 \of {\frac{{x - 1}}{{x + 1}}} \left( {\frac{1}{{x - 1}} - \frac{1}{{x + 1}}} \right) \cr
& \frac{{dy}}{{dx}} = \frac{1}{5}\root 5 \of {\frac{{x - 1}}{{x + 1}}} \left( {\frac{2}{{\left( {x - 1} \right)\left( {x + 1} \right)}}} \right) \cr
& \frac{{dy}}{{dx}} = \frac{2}{{5\left( {{x^2} - 1} \right)}}\root 5 \of {\frac{{x - 1}}{{x + 1}}} \cr} $$
